What have you observed on the two table?

Suppose that 10 sophomores, 3 juniors, and 10 seniors are candidates for a prestigious mathematics award of which three will be

faltersainse [42]

Answer:

1771 possible ways

Step-by-step explanation:

In this case, we need to know first how many candidates are in total:

10 + 3 + 10 = 23 candidates in total.

Now, we need to choose 3 of them to receive an award. In this case, we have several scenarios, but as it's an award we can also assume that the order in which the candidates are chosen do not matter, so, the formula to use is the following:

C = m! / n! (m - n)!

Where m is the total candidates and n, is the number of candidates to be chosen. Replacing this data we have:

C = 23! / 3! (23 - 3)!

C = 2.59x10^22 / 6(2.43x10^18)

C = 1771

So we have 1771 ways of choose the candidates.

7 0

2 years ago

Teresa's restaurant bill comes to $29.99 before tax.If the sales tax is 6.25% and she tips the waiter 20%, what is the total cos

Andre45 [30]

The total is $29.99. Tip is $5.98. Tax is $1.87. Total is $ 37.84.

3 0

2 years ago

Solve the following using Substitution method<br> 2x – 5y = -13<br><br> 3x + 4y = 15

Digiron [165]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

$2x=5y-13$

Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.